Let $J$ be an almost complex structure on an algebraic variety $V$. As we all know, $J$ comes from a complex structure if the Nijenhuis tensor of $J$ vanishes. What I would like to know is if there exists a simpler characterisation of integrability than this for varieties (as opposed to general manifolds).

When you say that $V$ is an algebraic variety, are you assuming that it is complex-algebraic (i.e., already has a complex structure), and that $J$ is an additional almost-complex structure?
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S. Carnahan♦Feb 21 '11 at 16:48

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@Janos: If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6
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Gunnar Þór MagnússonFeb 21 '11 at 18:03

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No, wait, do you want to hit two birds with one stone and get integrability for the almost complex structure and projectivity of the resulting complex manifold in one swoop? I'm not sure that's going to be doable... just look at two-dimensional tori. You can realize any complex torus as an integrable complex structure on $M = \mathbb R^2/\mathbb Z^2$, but not all tori are projective. Given an arbitrary integrable complex structure on $M$, I don't know how to link its properties to the question of projectivity of the torus.
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Gunnar Þór MagnússonFeb 21 '11 at 18:10

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There is a simpler criterion that does integrability and Kählerness in one step: if there is both a Riemann metric $g$ and an almost complex structure $J$ and $\nabla J =0$ (i.e., $J$ is parallel w.r.t. the Levi-Civita connection of the Riemann metric), then $J$ is automatically integrable and $g$ is a kähler metric.
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Johannes EbertFeb 23 '11 at 22:13

1 Answer
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If you want equivalent conditions to the Nijenhuis tensor vanishing then one is that the induced $\bar \partial$ operator defines a complex, i.e. that $\bar \partial^2 = 0$. Another one is that the exterior derivative decomposes as $d = \partial + \bar \partial$. If you want explicit examples of almost complex manifolds that are not complex, that's going to be more difficult, see the answers to tinyurl.com/4zrkar6

To find the $\bar \partial$ operator associated to an almost complex structure $J$ on a smooth manifold $M$, one needs to note that $J$ induces a splitting $T_M \otimes \mathbb C = T^{1,0} \oplus T^{0,1}$ of the tangent bundle into $i$ and $-i$ eigenvectors (the eigenspaces being marked with $(1,0)$ and $(0,1)$, respectively).

The same thing happens on the level of 1-forms (and indeed on the level of $k$-forms): they split into $(p,q)$-forms like on complex manifolds. If $\pi^{p,q} : \bigwedge^k T_M \to \bigwedge^{p,q} T_M$ is the projection onto the space of $(p,q)$-forms, then the $\bar \partial : \bigwedge^{p,q} T_M \to \bigwedge^{p,q+1} T_M$ operator associated to $J$ is $\bar \partial_J = \pi^{p,q+1} \circ d$.

Once one does the calculations this comes out to
$$ \bar \partial \alpha = \frac 1 2 \left( d \alpha + i d J \alpha \right) $$
for a $(p,q)$-form $\alpha$. A similar formula holds for the $\partial$ operator, you just have to change $i$ to $-i$. (I may have confounded the signs here.)

A very good reference for the linear algebra parts (i.e. most of this) is Chapter 2 of Huybrecht's "Complex geometry".

For the conditions equivalent to the vanishing of the Nijenhuis tensor I seem to remember the first chapter of http://tinyurl.com/4cqspu7 Moroianu's notes on Kahler geometry being very helpful when I went through this a couple of months ago.